∠B=∠C=90°,E是BC的中点,EF垂直AD,DE平分∠ADC,求证AE平分∠DAB

let me tell you——过E作EF⊥AD交DA于F∵DE平分∠ADC∴∠CDE=∠FDE在△EDF和△EDC中｛∠D=∠DFE∠FDE=∠CDEDE=DE ∴△EDF≌△EDC（AAS）∴FE=CE∵E平分BC∴CE=EB∴FE=EB在RT△FAE和RT△BAE中｛AE=AEFE=EB∴RT△FAE...